Question

___100 points___

41 tuning forks are arranged in a series in such a way that each T. F. produce 3 beats with its neighbor T. F. if the frequency of last is 3 times of first then find the frequency of 1st ,11th, 16th , 21st and last T. F. ?​

Answer 1

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⇒60 Hz, 69 Hz, 105 Hz, 120 Hz, 180 Hz

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ɢɪᴠᴇɴ–

Number of tunning fork = 41

Beats produced = 3

Last frequency = 3 times the first

ᴛᴏ ғɪɴᴅ–

Frequncy of 1st ,11th, 16th , 21st and last T. F =?

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ᴇxᴘʟᴀɴᴀᴛɪᴏɴ–

Let be the first frequncy be n

Then the successive frequency will be

n, n+3, n+6....3n

3n means as last is 3 times the first

Now we can see that above series are in AP

So we will apply AP last term formula

That is

L = A + (N-1)d

Here,

L is the last term

A is the first 

N no of terms

d is the difference

Now, putting the value

⇒3n = n + (41 - 1)3

⇒2n = 120

⇒n = 60

Therefore The above series will look like

60, 63, 66.....180

Now,

1 st ⇒ 60 Hz

11 th ⇒60 + 9 = 69 Hz

16 th⇒ 60 + 45 = 105 Hz

21 th⇒60 + 60 = 120 Hz

Last ⇒180 Hz

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ᴛʜᴇʀᴇғᴏʀᴇ, 1st ,11th, 16th , 21st and last T. F. are 60 Hz, 69 Hz, 105 Hz, 120 Hz, 180 Hz

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Answer 2

41 frequencies are in AP with common difference = 3

f(l) = 3f(1)=f(1)+(41-1)*3

2f(1) = 120

f(1) = 60 Hz

f(11) = 60 + (10)*3 = 90 Hz

f(16) = 60 + (15)*3 = 105

f(21) = 60 + (20)*3 = 120 Hz

f(l)= 3*f(1) = 3*60 = 180 Hz